Question

Find a?+b2+c2 if a+b+c = 20 and ab+bc+ca=71.​

Answer 1

Answer:

(a + b + c)2 = a2  +b2 + c2 + 2ab + 2bc + 2ca

⇒ (a + b + c)2 = 83 + 2(ab + bc + ca)

⇒ (a + b + c)2 = 83 + 2 × 71

⇒ (a + b + c)2 = 83 + 142

⇒ (a + b + c)2 = 225

⇒ a + b + c =

⇒ a + b + c = 15

Step-by-step explanation:

Answer 2

FINAL ANSWER:-

a² + b² + c² = 260

or

a = $4\sqrt{15 - a - b}$

GIVEN:-

a+b+c = 20

and

ab+bc+ca=71.​

TO FIND:-

Find a?

FORMULA USED:-

(a + b +c)² = a² + b² + c² + 2(ab + bc + ca)

VALUES:-

taking a2+b2+c2 as x

SOLUTION:-

(a + b +c)² = a² + b² + c² + 2(ab + bc + ca)

(20)² = a² + b² + c² + 2(70)

400 = a² + b² + c² + 140

a² + b² + c² = 400 - 140

a² + b² + c² = 260

or

a² = 260 - b² - c²

a = $\sqrt{260 - b^2 - c^2}$

a = $4\sqrt{15 - a - b}$

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